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JEE Advanced2025PhysicsOscillationsActual

The center of a disk of radius r and mass m is attached to a spring of spring constant k , inside a ring of radius R>r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery

Options

  1. A2 3 ( g R-r + k m )
  2. B2 g 3(R-r) + k m
  3. C1 6 ( g R-r + k m )
  4. D1 4 ( g R-r + k m )

Correct answer

A. 2 3 ( g R-r + k m )

Step-by-step solution

E= 1 2 k(R-r)^2 ^2+m g(R-r)(1- )+ 1 2 m v^2+ 1 2 m r^2 2 ^2 Differentiating wrt t , aligned 0 & = 1 2 k ( R - r )^2 2 ~d dt + mg ( R - r ) d dt (2 ^2 4 )+ 1 2 ~m 2 v dv dt + mr ^2 4 2 ~d dt & 0= k ( R - r )^2 ~d dt + mg ( R - r ) d dt + mv dv dt + mr ^2 2 ~d dt aligned Also, d d t = V (R-r) d^2 d t^2 = 1 (R-r) d v d t = 1 R-r a aligned & k ( R - r )^2 ~V R - r + mg ( R - r ) V R - r =- mv r - mr ^2 2 ~V r & k ( R - r )+ mg =- 3 2 mr & -[ k ( R - r )+ mg ] = 3 2 ~m ( R - r ) d ^2 dt ^2 & - 2 3 [ k m + g R - r ]= d ^

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